For a representable morphism of algebraic spaces we have already defined (in Section 67.3) what it means to be universally open. Hence before we give the natural definition we check that it agrees with this in the representable case.
Lemma 67.6.1. Let $S$ be a scheme. Let $f : X \to Y$ be a representable morphism of algebraic spaces over $S$. The following are equivalent
$f$ is universally open (in the sense of Section 67.3), and
for every morphism of algebraic spaces $Z \to Y$ the morphism of topological spaces $|Z \times _ Y X| \to |Z|$ is open.
Proof. Assume (1), and let $Z \to Y$ be as in (2). Choose a scheme $V$ and a surjective étale morphism $V \to Y$. By assumption the morphism of schemes $V \times _ Y X \to V$ is universally open. By Properties of Spaces, Section 66.4 in the commutative diagram
the horizontal arrows are open and surjective, and moreover
is surjective. Hence as the left vertical arrow is open it follows that the right vertical arrow is open. This proves (2). The implication (2) $\Rightarrow $ (1) is immediate from the definitions. $\square$
Thus we may use the following natural definition.
Definition 67.6.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.
We say $f$ is open if the map of topological spaces $|f| : |X| \to |Y|$ is open.
We say $f$ is universally open if for every morphism of algebraic spaces $Z \to Y$ the morphism of topological spaces
is open, i.e., the base change $Z \times _ Y X \to Z$ is open.
\[ |Z \times _ Y X| \to |Z| \]
Note that an étale morphism of algebraic spaces is universally open, see Properties of Spaces, Definition 66.16.2 and Lemmas 66.16.7 and 66.16.5.
Lemma 67.6.3. The base change of a universally open morphism of algebraic spaces by any morphism of algebraic spaces is universally open.
Proof. This is immediate from the definition. $\square$
Lemma 67.6.4. The composition of a pair of (universally) open morphisms of algebraic spaces is (universally) open.
Proof. Omitted. $\square$
Lemma 67.6.5. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent
$f$ is universally open,
for every scheme $Z$ and every morphism $Z \to Y$ the projection $|Z \times _ Y X| \to |Z|$ is open,
for every affine scheme $Z$ and every morphism $Z \to Y$ the projection $|Z \times _ Y X| \to |Z|$ is open, and
there exists a scheme $V$ and a surjective étale morphism $V \to Y$ such that $V \times _ Y X \to V$ is a universally open morphism of algebraic spaces, and
there exists a Zariski covering $Y = \bigcup Y_ i$ such that each of the morphisms $f^{-1}(Y_ i) \to Y_ i$ is universally open.
Proof. We omit the proof that (1) implies (2), and that (2) implies (3).
Assume (3). Choose a surjective étale morphism $V \to Y$. We are going to show that $V \times _ Y X \to V$ is a universally open morphism of algebraic spaces. Let $Z \to V$ be a morphism from an algebraic space to $V$. Let $W \to Z$ be a surjective étale morphism where $W = \coprod W_ i$ is a disjoint union of affine schemes, see Properties of Spaces, Lemma 66.6.1. Then we have the following commutative diagram
We have to show the south-east arrow is open. The middle horizontal arrows are surjective and open (Properties of Spaces, Lemma 66.16.7). By assumption (3), and the fact that $W_ i$ is affine we see that the left vertical arrows are open. Hence it follows that the right vertical arrow is open.
Assume $V \to Y$ is as in (4). We will show that $f$ is universally open. Let $Z \to Y$ be a morphism of algebraic spaces. Consider the diagram
The south-west arrow is open by assumption. The horizontal arrows are surjective and open because the corresponding morphisms of algebraic spaces are étale (see Properties of Spaces, Lemma 66.16.7). It follows that the right vertical arrow is open.
Of course (1) implies (5) by taking the covering $Y = Y$. Assume $Y = \bigcup Y_ i$ is as in (5). Then for any $Z \to Y$ we get a corresponding Zariski covering $Z = \bigcup Z_ i$ such that the base change of $f$ to $Z_ i$ is open. By a simple topological argument this implies that $Z \times _ Y X \to Z$ is open. Hence (1) holds. $\square$
Lemma 67.6.6. Let $S$ be a scheme. Let $p : X \to \mathop{\mathrm{Spec}}(k)$ be a morphism of algebraic spaces over $S$ where $k$ is a field. Then $p : X \to \mathop{\mathrm{Spec}}(k)$ is universally open.
Proof. Choose a scheme $U$ and a surjective étale morphism $U \to X$. The composition $U \to \mathop{\mathrm{Spec}}(k)$ is universally open (as a morphism of schemes) by Morphisms, Lemma 29.23.4. Let $Z \to \mathop{\mathrm{Spec}}(k)$ be a morphism of schemes. Then $U \times _{\mathop{\mathrm{Spec}}(k)} Z \to X \times _{\mathop{\mathrm{Spec}}(k)} Z$ is surjective, see Lemma 67.5.5. Hence the first of the maps
is surjective. Since the composition is open by the above we conclude that the second map is open as well. Whence $p$ is universally open by Lemma 67.6.5. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)